Six Sigma Green Belt Examination

July 30, 2017 | Author: Navneet | Category: Six Sigma, Normal Distribution, Standard Deviation, Statistics, Analysis

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Descripción: Question Paper of Six Sigma Green Belt Examination...

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SIX SIGMA GREEN BELT EXAMINATION 1. The mentor for the Green Belt should at least be a (A) Master Black belt (B) Champion (C) Black Belt (D) Green Belt Ans: C 2. The Black Belt is guiding projects carried out by green or yellow belts on a (A) Full time basis (B) Partial basis (C) Beyond official working hours (D) Consulting basis Ans:

A

3. The data for projects is collected by the (A) Green Belts (B) Black belts (C) Both of them (D) None of them A

Ans:

4. Metric of SIX SIGMA gives an insight into how many (A) Defectives (B) Defects (C) Defects per million opportunity (D) Defects per unit Ans:

C

5. SIX SIGMA concepts were born in (A) Allied Signal Ltd. (B) General Electric Ltd. (C) Chrysler Ltd. (D) Motorola Ltd. Ans:

D

6. The person who propagated SIX SIGMA concepts in General Electric Ltd. .was : (A) Jack Welch (B) Bob Smith (C) Philip Crosby (D) None of them

Ans:

A

7. Weaving of Cloth is being done in looms. It was seen that defects like coagulation of thread, Knot, Breakage in continuity of thread, Grease marks are opportunities of defects. The number of such defects were found to be 20 in the 100 dhotis woven. Calculate the six sigma level at which the looms was operating. Ans: No. of Defects = 20 Total No. of Dhotis Woven = 100 No. of Opportunities = 4 (coagulation of threads, knot, breakage of continuity of threads & grease marks) Defects per unit (DPU) = D/U = 20/100 = 0.20 Defects per opportunities (DPO) = 20/ (4*100) = 0.05 Defects per million opportunities (DPMO) = 0.05*106 = 50000 PPM Z = 0.5 – 0.05 = 0.45 Z Value = 1.645 Sigma Value to Z = 1.645 + 1.5 = 3.145 8. Kano Model shows the path towards (A) Satisfying a customer (B) Delighting a customer (C) Ensure basic needs of a customer (D) None of the above Ans:

B

9. Pareto Chart is read by the user to find out (A) which 20% of the causes contribute towards 80% of the problem (B) which 20% of the people in a state posses 80% of the wealth (C) Both of the above (D) None of the above. Ans:

C

10. Cause and Effect Diagram was first formulated by (A) Juran (B) Deming (C) Taguchi (D) Ishikawa Ans:

D

11. In Measurement System Analysis the variance contribution allowed for Gage R & R is to a maximum of (A) 20% (B) 10%

(C) 30% (D) 5% Ans:

C

12. The following data on an experiment is given. Perform the measurement System Analysis on the data. Parts 1 2 3 4 5 Ans: Parts

Operator 1 Trial 1 10.2 11.3 10.9 14.0 12.0 11.6 12.6 15.0 10.2 10.3

Trial 2 11.6 11.7 11.8 11.5 10.2 10.6 10.5 14.2 14.6 12.3

Operator 2 Trial 1 11.0 11.2 11.0 11.3 11.2 10.3 11.4 11.2 11.3 11.6

Trial 2 11.2 12.5 12.8 12.7 13.0 13.4 10.3 14.3 13.2 10.8

Operator 1 Operator 2 Trial 1 Trial 2 avg range Trial 1 Trial 2 avg range 1 10.2 11.6 11 11.2 11.3 11.7 11.2 12.5 Avg 10.75 11.65 11.2 11.1 11.85 11.475 range 1.1 0.1 0.6 0.2 1.3 0.75 2 10.9 11.8 11 12.8 14 11.5 11.3 12.7 Avg 12.45 11.65 12.05 11.15 12.75 11.95 range 3.1 0.3 1.7 0.3 1.3 0.8 3 12 10.2 11.2 13 11.6 10.6 10.3 13.4 Avg 11.8 10.4 11.1 10.75 13.2 11.975 range 0.4 0.4 0.4 0.9 0.4 0.65 4 12.6 10.5 11.4 10.3 15 14.2 11.2 14.3 Avg 13.8 12.35 13.075 11.3 12.3 11.8 range 2.4 3.7 3.05 -0.2 4 1.9 5 10.2 14.6 11.3 13.2 10.3 12.3 11.6 10.8 Avg 11.85 10.25 13.45 11.45 12 11.725 range 1.2 1.35 0.1 2.3 0.3 2.4 Mean of Xa=11.855;

Mean of Xb=11.785; Range of A=1.39; Range of B=1.09 X diff=0.07 13. The normality check on the data is done through normal probability paper where the Anderson Darling test is performed. For the data to be normal the p-value should be (A) 0.05 (B) >0.05 (C)